Question: A particle moves along the curve $y=3x^3-30$ so that the $x$ -coordinate is increasing at a constant rate of $3$ units per second. What is the rate of change (in units per second) of the particle's $y$ -coordinate when the particle is at the point $(3,51)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $51$ (Choice B) B $27$ (Choice C) C $-30$ (Choice D) D $243$
Background When working with motion along a curve, we should remember that this motion can also be represented by a position vector $(x,y)$. The main difference is that we're not given the equations for $x$ and $y$ in terms of time $t$, but only the relationship between $x$ and $y$ themselves. This, however, shouldn't prevent us from assuming the expressions for $x$ and $y$ in terms of $t$ exist. As always, $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$. Setting up the math We are given that $\dfrac{dx}{dt}=3$ for any value of $t$. We are asked for the rate of change of the particle's $y$ -coordinate when the particle is at the point $(3,51)$. In other words, we are asked for the value of $\dfrac{dy}{dt}$ at the point $(3,51)$. Finding $\dfrac{dy}{dt}$ $\dfrac{dy}{dt}=27x^2$ Finding $\dfrac{dy}{dt}$ at $(3,51)$ The expression for $\dfrac{dy}{dt}$ only depends on the particle's $x$ -coordinate, which in our case is ${x}={3}$ : $\begin{aligned} \dfrac{dy}{dt}&=27({3})^2 \\\\ &=243 \end{aligned}$ In conclusion, the rate of change of the particle's $y$ -coordinate when the particle is at the point $(3,51)$ is $243$ units per second.